I am trying to use if-else expression which is supposed to break the loop if the if condition fails, but getting an invalid syntax error.
Sample code:
a = 5
while True:
print(a) if a > 0 else break
a-=1
Of course, if I write in the traditional way (not using the one liner) it works.
What is wrong in using the break command after the else keyword?
If I run this, I get the following error:
... print(a) if a > 0 else break
File "<stdin>", line 2
print(a) if a > 0 else break
^
SyntaxError: invalid syntax
This is because
print(a) if a > 5 else break
is a ternary operator. Ternary operators are no if statements. These work with syntax:
<expr1> if <expr2> else <expr3>
It is equivalent to a "virtual function":
def f():
if <expr2>:
return <expr1>
else:
return <expr3>
So that means the part next to the else should be an expression. break is not an expression, it is a statement. So Python does not expect that. You can not return a break.
In python-2.x, print was not a function either. So this would error with the print statement. In python-2.x print was a keyword.
You can rewrite your code to:
a = 5
while True:
if a > 5:
print(a)
else:
break
a -= 1
You can read more about this in the documentation and PEP-308.
If is an expression, break similar to return is a statement. You can't use two statements in a single sentence (unless you use a semicolon which is ugly). I know it would have been really cool if we can do that, but alas that's the way it is.
To put it in slightly simpler terms, you're misusing the 'one-line if statement' (ternary operator). It always evaluates to an expression (i.e., a value). That is,
<expr1> if <condition> else <expr2>
evaluates to <expr1> if <condition> is True, and to <expr2> if <condition> is False. This resulting value can then be used like any Python value, for example:
y = 0
x = (5 if (y > 0) else 6)
print(x) # 6
Of course, the parentheses are completely unnecessary (even discouraged), but hopefully are useful for understanding the meaning of that line.
Therefore,
print(a) if a > 0 else break
tries to evaluate print(a) (which, by the definition of print() in Python 3, always returns None – perfectly valid, but probably not what you usually want) and then break, which does not evaluate to anything because it is a statement (action), not an expression (value), hence the invalid syntax error.
Hence, if you want to execute one of two statements depending on a condition, you really need the multi-line solution proposed by
Willem Van Onsem. There may be hacky ways to do it in one line, but multiple lines is the usual solution for something like this in Python.
I have a small snippet of code with two functions in it.
I want to call the first function if it receives a response then perform a function on that response. Then assign the result to another variable.
In a verbose way it looks like:
result = get_something()
if result:
answer = transform(result)
alternatively I could do
if get_something():
answer = transform(get_something())
but that requires calling the first function twice
is there a way to do all of this on one line a bit like a ternary (maybe as a lambda)
answer = transform(result) if get_something() else None
Obviously in the above there is nothing to state what result is but I need to say basically where result = get_something()
I can do that in a list comprehension but that seems a bit dumb
answer = [transform(x) for x in [get_something()] if x][0]
In the latest Python version (Python 3.8) there's a new assignment that may be useful for you, :=:
There is new syntax := that assigns values to variables as part of a larger expression. It is affectionately known as “walrus operator” due to its resemblance to the eyes and tusks of a walrus.
if (n := len(a)) > 10:
print(f"List is too long ({n} elements, expected <= 10)")
In this example, the assignment expression helps avoid calling len() twice:
We can in Python 3.8 with assignment expressions:
if (result := get_something()) is not None:
# do something with result
Although I don't fully understand the reasons for doing things this way (which is less clear than any of the others), here's an example using lambda:
>>> def get_something(flag): # Added the flag argument, to mimic different return values
... return 5 if flag else None
...
>>> answer = (lambda func, arg: func(arg) if arg else None)(int, get_something(True))
>>> answer
5
>>> answer = (lambda func, arg: func(arg) if arg else None)(int, get_something(False))
>>> answer
>>>
I would like to have a function AllTrue that takes three arguments:
List: a list of values
Function: a function to apply to all values
Condition: something to test against the function's output
and return a boolean of whether or not all values in the list match the criteria.
I can get this to work for basic conditions as follows:
def AllTrue(List, Function = "Boolean", Condition = True):
flag = True
condition = Condition
if Function == "Boolean"
for element in List:
if element != condition:
flag = False
break
else:
Map = map(Function, List)
for m in Map:
if m != condition:
flag = False
break
return flag
Since python doesn't have function meant for explicitly returning if something is True, I just make the default "Boolean". One could clean this up by defining TrueQ to return True if an element is True and then just mapping TrueQ on the List.
The else handles queries like:
l = [[0,1], [2,3,4,5], [6,7], [8,9],[10]]
AllTrue(l, len, 2)
#False
testing if all elements in the list are of length 2. However, it can't handle more complex conditions like >/< or compound conditions like len > 2 and element[0] == 15
How can one do this?
Cleaned up version
def TrueQ(item):
return item == True
def AllTrue(List, Function = TrueQ, Condition = True):
flag = True
condition = Condition
Map = map(Function, List)
for m in Map:
if m != condition:
flag = False
break
return flag
and then just call AllTrue(List,TrueQ)
Python already has built-in the machinery you are trying to build. For example to check if all numbers in a list are even the code could be:
if all(x%2==0 for x in L):
...
if you want to check that all values are "truthy" the code is even simpler:
if all(L):
...
Note that in the first version the code is also "short-circuited", in other words the evaluation stops as soon as the result is known. In:
if all(price(x) > 100 for x in stocks):
...
the function price will be called until the first stock is found with a lower or equal price value. At that point the search will stop because the result is known to be False.
To check that all lengths are 2 in the list L the code is simply:
if all(len(x) == 2 for x in L):
...
i.e. more or less a literal translation of the request. No need to write a function for that.
If this kind of test is a "filter" that you want to pass as a parameter to another function then a lambda may turn out useful:
def search_DB(test):
for record in database:
if test(record):
result.append(record)
...
search_DB(lambda rec: all(len(x) == 2 for x in rec.strings))
I want a function that takes a list, a function, and a condition, and tells me if every element in the list matches the condition. i.e. foo(List, Len, >2)
In Python >2 is written lambda x : x>2.
There is (unfortunately) no metaprogramming facility in Python that would allow to write just >2 or things like ·>2 except using a string literal evaluation with eval and you don't want to do that. Even the standard Python library tried going down that path (see namedtuple implementation in collections) but it's really ugly.
I'm not saying that writing >2 would be a good idea, but that it would be nice to have a way to do that in case it was a good idea. Unfortunately to have decent metaprogramming abilities you need a homoiconic language representing code as data and therefore you would be programming in Lisp or another meta-language, not Python (programming in Lisp would indeed be a good idea, but for reasons unknown to me that approach is still unpopular).
Given that, the function foo to be called like
foo(L, len, lambda x : x > 2)
is just
def foo(L, f=lambda x : x, condition=lambda x: x):
return all(condition(f(x)) for x in L)
but no Python programmer would write such a function, because the original call to foo is actually more code and less clear than inlining it with:
all(len(x) > 2 for x in L)
and requires you to also learn about this thing foo (that does what all and a generator expression would do, just slower, with more code and more obfuscated).
You are reinventing the wheel. Just use something like this:
>>> l = [[0,1], [2,3,4,5], [6,7], [8,9],[10]]
>>> def all_true(iterable, f, condition):
... return all(condition(f(e)) for e in iterable)
...
>>> def cond(x): return x == 2
...
>>> all_true(l, len, cond)
False
You can define a different function to check a different condition:
>>> def cond(x): return x >= 1
...
>>> all_true(l, len, b)
True
>>>
And really, having your own function that does this seems like overkill. For example, to deal with your "complex condition" you could simply do something like:
>>> l = [[0,2],[0,1,2],[0,1,3,4]]
>>> all(len(sub) > 2 and sub[0] == 5 for sub in l)
False
>>> all(len(sub) > 1 and sub[0] == 0 for sub in l)
True
>>>
I think the ideal solution in this case may be:
def AllTrue(List, Test = lambda x:x):
all(Test(x) for x in List)
This thereby allows complex queries like:
l = [[0, 1], [1, 2, 3], [2, 5]]
AllTrue(l, lambda x: len(x) > 2 and x[0] == 1)
To adhere to Juanpa's suggestion, here it is in python naming conventions and an extension of what I posted in the question now with the ability to handle simple conditions like x > value.
from operator import *
all_true(a_list, a_function, an_operator, a_value):
a_map = map(a_function, a_list)
return all( an_operator(m, a_value) for m in a_map)
l = [[0,2],[0,1,2],[0,1,3,4]]
all_true(l, len, gt, 2)
#True
Note: this works for single conditions, but not for complex conditions like
len > 2 and element[0] == 5
I was going over an assignment, and came across something that confused me, as am I not crazy good with python. Here is the code.
def main():
list = [1,2]
x = 2
if (x in list == True):
print("hi")
if (x in list):
print("Why does this proc?")
main()
I believed the output would be both, but the output is only the second if statement. I know that in C, if you had something like
if (x = 6)
That since there is only one '=' that x is now equal to 6. (As its read, if (), x = 6).
Is something similar happening to this python code? Is it checking 'list == true' first, then from there checking about x being in list?
Any insight would be greatly appreciated!
As you can see, yes, your expression requires explicit grouping:
>>> 2 in [1,2] == True
False
>>> (2 in [1,2]) == True
True
Note that, as #tavo and #MorganThrapp mention, the version without parentheses is doing a chained comparison, checking that 2 in [1,2] and then checking that [1,2] == True. The latter is false, so the full expression is also false.
By the way, don't name your variables after built-ins like list, or you won't be able to use those functions easily.
Also, you don't have to compare the result of an expression to True:
>>> 2 in [1,2]
True
Doing so is the equivalent of asking "is 'the cake is ready' a true statement?" as opposed to "is the cake ready?".
I'm calling a bunch of methods that return a list. The list may be empty. If the list is non-empty, I want to return the first item; otherwise, I want to return None. This code works:
def main():
my_list = get_list()
if len(my_list) > 0:
return my_list[0]
return None
but it seems to me that there should be a simple one-line idiom for doing this. Is there?
Python 2.6+
next(iter(your_list), None)
If your_list can be None:
next(iter(your_list or []), None)
Python 2.4
def get_first(iterable, default=None):
if iterable:
for item in iterable:
return item
return default
Example:
x = get_first(get_first_list())
if x:
...
y = get_first(get_second_list())
if y:
...
Another option is to inline the above function:
for x in get_first_list() or []:
# process x
break # process at most one item
for y in get_second_list() or []:
# process y
break
To avoid break you could write:
for x in yield_first(get_first_list()):
x # process x
for y in yield_first(get_second_list()):
y # process y
Where:
def yield_first(iterable):
for item in iterable or []:
yield item
return
The best way is this:
a = get_list()
return a[0] if a else None
You could also do it in one line, but it's much harder for the programmer to read:
return (get_list()[:1] or [None])[0]
(get_list() or [None])[0]
That should work.
BTW I didn't use the variable list, because that overwrites the builtin list() function.
The most python idiomatic way is to use the next() on a iterator since list is iterable. just like what #J.F.Sebastian put in the comment on Dec 13, 2011.
next(iter(the_list), None) This returns None if the_list is empty. see next() Python 2.6+
or if you know for sure the_list is not empty:
iter(the_list).next() see iterator.next() Python 2.2+
If you find yourself trying to pluck the first thing (or None) from a list comprehension you can switch to a generator to do it like:
next((x for x in blah if cond), None)
Pro: works if blah isn't indexable Con: it's unfamiliar syntax. It's useful while hacking around and filtering stuff in ipython though.
The OP's solution is nearly there, there are just a few things to make it more Pythonic.
For one, there's no need to get the length of the list. Empty lists in Python evaluate to False in an if check. Just simply say
if list:
Additionally, it's a very Bad Idea to assign to variables that overlap with reserved words. "list" is a reserved word in Python.
So let's change that to
some_list = get_list()
if some_list:
A really important point that a lot of solutions here miss is that all Python functions/methods return None by default. Try the following below.
def does_nothing():
pass
foo = does_nothing()
print foo
Unless you need to return None to terminate a function early, it's unnecessary to explicitly return None. Quite succinctly, just return the first entry, should it exist.
some_list = get_list()
if some_list:
return list[0]
And finally, perhaps this was implied, but just to be explicit (because explicit is better than implicit), you should not have your function get the list from another function; just pass it in as a parameter. So, the final result would be
def get_first_item(some_list):
if some_list:
return list[0]
my_list = get_list()
first_item = get_first_item(my_list)
As I said, the OP was nearly there, and just a few touches give it the Python flavor you're looking for.
Python idiom to return first item or None?
The most Pythonic approach is what the most upvoted answer demonstrated, and it was the first thing to come to my mind when I read the question. Here's how to use it, first if the possibly empty list is passed into a function:
def get_first(l):
return l[0] if l else None
And if the list is returned from a get_list function:
l = get_list()
return l[0] if l else None
New in Python 3.8, Assignment Expressions
Assignment expressions use the in-place assignment operator (informally called the walrus operator), :=, new in Python 3.8, allows us to do the check and assignment in-place, allowing the one-liner:
return l[0] if (l := get_list()) else None
As a long-time Python user, this feels like we're trying to do too much on one line - I feel it would be better style to do the presumptively equally performant:
if l := get_list():
return l[0]
return None
In support of this formulation is Tim Peter's essay in the PEP proposing this change to the language. He didn't address the first formulation, but based on the other formulations he did like, I don't think he would mind.
Other ways demonstrated to do this here, with explanations
for
When I began trying to think of clever ways to do this, this is the second thing I thought of:
for item in get_list():
return item
This presumes the function ends here, implicitly returning None if get_list returns an empty list. The below explicit code is exactly equivalent:
for item in get_list():
return item
return None
if some_list
The following was also proposed (I corrected the incorrect variable name) which also uses the implicit None. This would be preferable to the above, as it uses the logical check instead of an iteration that may not happen. This should be easier to understand immediately what is happening. But if we're writing for readability and maintainability, we should also add the explicit return None at the end:
some_list = get_list()
if some_list:
return some_list[0]
slice or [None] and select zeroth index
This one is also in the most up-voted answer:
return (get_list()[:1] or [None])[0]
The slice is unnecessary, and creates an extra one-item list in memory. The following should be more performant. To explain, or returns the second element if the first is False in a boolean context, so if get_list returns an empty list, the expression contained in the parentheses will return a list with 'None', which will then be accessed by the 0 index:
return (get_list() or [None])[0]
The next one uses the fact that and returns the second item if the first is True in a boolean context, and since it references my_list twice, it is no better than the ternary expression (and technically not a one-liner):
my_list = get_list()
return (my_list and my_list[0]) or None
next
Then we have the following clever use of the builtin next and iter
return next(iter(get_list()), None)
To explain, iter returns an iterator with a .next method. (.__next__ in Python 3.) Then the builtin next calls that .next method, and if the iterator is exhausted, returns the default we give, None.
redundant ternary expression (a if b else c) and circling back
The below was proposed, but the inverse would be preferable, as logic is usually better understood in the positive instead of the negative. Since get_list is called twice, unless the result is memoized in some way, this would perform poorly:
return None if not get_list() else get_list()[0]
The better inverse:
return get_list()[0] if get_list() else None
Even better, use a local variable so that get_list is only called one time, and you have the recommended Pythonic solution first discussed:
l = get_list()
return l[0] if l else None
Regarding idioms, there is an itertools recipe called nth.
From itertools recipes:
def nth(iterable, n, default=None):
"Returns the nth item or a default value"
return next(islice(iterable, n, None), default)
If you want one-liners, consider installing a library that implements this recipe for you, e.g. more_itertools:
import more_itertools as mit
mit.nth([3, 2, 1], 0)
# 3
mit.nth([], 0) # default is `None`
# None
Another tool is available that only returns the first item, called more_itertools.first.
mit.first([3, 2, 1])
# 3
mit.first([], default=None)
# None
These itertools scale generically for any iterable, not only for lists.
for item in get_list():
return item
Frankly speaking, I do not think there is a better idiom: your is clear and terse - no need for anything "better". Maybe, but this is really a matter of taste, you could change if len(list) > 0: with if list: - an empty list will always evaluate to False.
On a related note, Python is not Perl (no pun intended!), you do not have to get the coolest code possible.
Actually, the worst code I have seen in Python, was also very cool :-) and completely unmaintainable.
By the way, most of the solution I have seen here do not take into consideration when list[0] evaluates to False (e.g. empty string, or zero) - in this case, they all return None and not the correct element.
my_list[0] if len(my_list) else None
Not sure how pythonic this is but until there is a first function in the library I include this in the source:
first = lambda l, default=None: next(iter(l or []), default)
It's just one line (conforms to black) and avoids dependencies.
Out of curiosity, I ran timings on two of the solutions. The solution which uses a return statement to prematurely end a for loop is slightly more costly on my machine with Python 2.5.1, I suspect this has to do with setting up the iterable.
import random
import timeit
def index_first_item(some_list):
if some_list:
return some_list[0]
def return_first_item(some_list):
for item in some_list:
return item
empty_lists = []
for i in range(10000):
empty_lists.append([])
assert empty_lists[0] is not empty_lists[1]
full_lists = []
for i in range(10000):
full_lists.append(list([random.random() for i in range(10)]))
mixed_lists = empty_lists[:50000] + full_lists[:50000]
random.shuffle(mixed_lists)
if __name__ == '__main__':
ENV = 'import firstitem'
test_data = ('empty_lists', 'full_lists', 'mixed_lists')
funcs = ('index_first_item', 'return_first_item')
for data in test_data:
print "%s:" % data
for func in funcs:
t = timeit.Timer('firstitem.%s(firstitem.%s)' % (
func, data), ENV)
times = t.repeat()
avg_time = sum(times) / len(times)
print " %s:" % func
for time in times:
print " %f seconds" % time
print " %f seconds avg." % avg_time
These are the timings I got:
empty_lists:
index_first_item:
0.748353 seconds
0.741086 seconds
0.741191 seconds
0.743543 seconds avg.
return_first_item:
0.785511 seconds
0.822178 seconds
0.782846 seconds
0.796845 seconds avg.
full_lists:
index_first_item:
0.762618 seconds
0.788040 seconds
0.786849 seconds
0.779169 seconds avg.
return_first_item:
0.802735 seconds
0.878706 seconds
0.808781 seconds
0.830074 seconds avg.
mixed_lists:
index_first_item:
0.791129 seconds
0.743526 seconds
0.744441 seconds
0.759699 seconds avg.
return_first_item:
0.784801 seconds
0.785146 seconds
0.840193 seconds
0.803380 seconds avg.
try:
return a[0]
except IndexError:
return None
def head(iterable):
try:
return iter(iterable).next()
except StopIteration:
return None
print head(xrange(42, 1000) # 42
print head([]) # None
BTW: I'd rework your general program flow into something like this:
lists = [
["first", "list"],
["second", "list"],
["third", "list"]
]
def do_something(element):
if not element:
return
else:
# do something
pass
for li in lists:
do_something(head(li))
(Avoiding repetition whenever possible)
Borrowing more_itertools.first_true code yields something decently readable:
def first_true(iterable, default=None, pred=None):
return next(filter(pred, iterable), default)
def get_first_non_default(items_list, default=None):
return first_true(items_list, default, pred=lambda x: x!=default)
Following code covers several scenarios by using lambda:
l1 = [1,2,3]
l2 = []
l3 = None
first_elem = lambda x: x[0] if x else None
print(first_elem(l1))
print(first_elem(l2))
print(first_elem(l3))
Using the and-or trick:
a = get_list()
return a and a[0] or None
Probably not the fastest solution, but nobody mentioned this option:
dict(enumerate(get_list())).get(0)
if get_list() can return None you can use:
dict(enumerate(get_list() or [])).get(0)
Advantages:
-one line
-you just call get_list() once
-easy to understand
My use case was only to set the value of a local variable.
Personally I found the try and except style cleaner to read
items = [10, 20]
try: first_item = items[0]
except IndexError: first_item = None
print first_item
than slicing a list.
items = [10, 20]
first_item = (items[:1] or [None, ])[0]
print first_item
How about this:
(my_list and my_list[0]) or None
Note: This should work fine for lists of objects but it might return incorrect answer in case of number or string list per the comments below.
You could use Extract Method. In other words extract that code into a method which you'd then call.
I wouldn't try to compress it much more, the one liners seem harder to read than the verbose version. And if you use Extract Method, it's a one liner ;)
Several people have suggested doing something like this:
list = get_list()
return list and list[0] or None
That works in many cases, but it will only work if list[0] is not equal to 0, False, or an empty string. If list[0] is 0, False, or an empty string, the method will incorrectly return None.
I've created this bug in my own code one too many times !
isn't the idiomatic python equivalent to C-style ternary operators
cond and true_expr or false_expr
ie.
list = get_list()
return list and list[0] or None
if mylist != []:
print(mylist[0])
else:
print(None)